Question

Tangents are drawn to a circle from a point which always lies on a given line; prove that the locus of the middle point of the chord of contact is another circle.

Answer 1

Let the given circle be $ax+by+c=0$ ....(1)

And the circle be $x^2+y^2=r^2$ ......(2)
Suppose $(\alpha ,\beta )$ is any point on $1$; then
$a\alpha +b\beta +c=0$ ....(3)
Chord of contact of $(\alpha ,\beta )$ w.r.t. $2$ is
$\alpha x+\beta y=r^2$
If $(h,k)$ be the middle point of this chord of contact $(h,k)$ will satisfy its equation
$\alpha h+\beta k=r^2$ ......(4)
Again if we draw a line perpendicular to the chord of contact through the center of the circle $(0,0)$, it will cut it at the middle point.
Equation of the line perpendicular to the chord of contact through the origin $(0,0)$ is $\beta h-\alpha k=0$ .....(5)
To get the locus, eliminate $\alpha$ and $\beta$ from $3,4$ and $5$
So solving $4$ and $5$ for $\alpha$ and $\beta$, we get
$\alpha =\frac{x{r}^2}{x^2+y^2}$ and $\beta =\frac{y{r}^2}{x^2+y^2}$
Putting these values in (3), we get
$a\frac{x{r}^2}{x^2+y^2}+b\frac{y{r}^2}{x^2+y^2}+c=0$
$\Rightarrow c{x}^2+x{y}^2+a{r}^{2}x+b{r}^{2}y=0$
which is a circle passing through origin